answer:To find the mass of silver deposited on the cathode, we can use the following steps: 1. Calculate the total charge passed through the solution. 2. Determine the number of moles of electrons transferred. 3. Calculate the number of moles of silver deposited. 4. Find the mass of silver deposited. Step 1: Calculate the total charge passed through the solution. Current (I) = 3.0 A Time (t) = 45 minutes = 45 × 60 = 2700 seconds Total charge (Q) = Current × Time Q = 3.0 A × 2700 s Q = 8100 C Step 2: Determine the number of moles of electrons transferred. Faraday's constant (F) = 96500 C/mol Moles of electrons (n_e) = Total charge / Faraday's constant n_e = 8100 C / 96500 C/mol n_e = 0.0839 mol Step 3: Calculate the number of moles of silver deposited. The reaction occurring at the cathode is: Ag⁺ + e⁻ → Ag From the balanced equation, we can see that 1 mole of silver ions (Ag⁺) requires 1 mole of electrons (e⁻) to be reduced to 1 mole of silver atoms (Ag). Therefore, the number of moles of silver deposited is equal to the number of moles of electrons transferred. Moles of silver (n_Ag) = Moles of electrons (n_e) n_Ag = 0.0839 mol Step 4: Find the mass of silver deposited. Molar mass of silver (M_Ag) = 107.87 g/mol Mass of silver (m_Ag) = Moles of silver × Molar mass of silver m_Ag = 0.0839 mol × 107.87 g/mol m_Ag = 9.05 g So, the mass of silver deposited on the cathode is 9.05 grams.

answer:To solve this problem, we will use Faraday's laws of electrolysis. The first law states that the mass of a substance deposited on an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. First, we need to find the total charge passed through the electrolyte. The charge (Q) can be calculated using the formula: Q = I × t where I is the current (in amperes) and t is the time (in seconds). Given, current (I) = 2.5 A and time (t) = 5 minutes = 5 × 60 = 300 seconds. Now, let's calculate the charge: Q = 2.5 A × 300 s = 750 C (coulombs) Next, we need to find the number of moles of electrons passed through the electrolyte. To do this, we will use the Faraday constant (F), which is the charge of one mole of electrons and is approximately 96,485 C/mol. Number of moles of electrons = Q / F Number of moles of electrons = 750 C / 96,485 C/mol ≈ 0.00777 mol Since the charge on each copper ion is 2+, it means that two moles of electrons are required to deposit one mole of copper. Therefore, the number of moles of copper deposited is half the number of moles of electrons: Number of moles of copper = 0.00777 mol / 2 = 0.003885 mol Now, we can find the mass of copper deposited using the atomic weight of copper (63.5 g/mol): Mass of copper = Number of moles of copper × Atomic weight of copper Mass of copper = 0.003885 mol × 63.5 g/mol ≈ 0.2466 g Therefore, the mass of copper deposited on the cathode is approximately 0.2466 grams.

answer:To find the mass of copper deposited, we can use Faraday's law of electrolysis. Faraday's law states that the mass of a substance deposited or dissolved at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. First, we need to find the charge passed through the electrolyte. Charge (Q) can be calculated using the formula: Q = I × t where I is the current (in amperes) and t is the time (in seconds). Given, current (I) = 2.5 A and time (t) = 25 minutes = 25 × 60 = 1500 seconds. Now, let's calculate the charge: Q = 2.5 A × 1500 s = 3750 Coulombs Next, we need to find the number of moles of electrons passed through the electrolyte. We can use the formula: n = Q / F where n is the number of moles of electrons, Q is the charge (in Coulombs), and F is Faraday's constant (approximately 96485 C/mol). n = 3750 C / 96485 C/mol ≈ 0.0388 mol The balanced half-reaction for the reduction of Cu^2+ ions to Cu metal is: Cu^2+ + 2e^- → Cu From the stoichiometry of the reaction, 2 moles of electrons are required to deposit 1 mole of copper. So, we can find the number of moles of copper deposited (n_Cu) using the formula: n_Cu = n / 2 n_Cu = 0.0388 mol / 2 ≈ 0.0194 mol Finally, we can find the mass of copper deposited (m_Cu) using the formula: m_Cu = n_Cu × M_Cu where M_Cu is the molar mass of copper (63.5 g/mol). m_Cu = 0.0194 mol × 63.5 g/mol ≈ 1.23 g Therefore, the mass of copper deposited is approximately 1.23 grams.

answer:First, we need to find the total charge passed through the solution during the electrolysis experiment. We can do this by multiplying the current (in amperes) by the time (in seconds). Current = 2 A Time = 30 minutes = 30 * 60 seconds = 1800 seconds Total charge = Current × Time Total charge = 2 A × 1800 s = 3600 Coulombs Next, we need to find the number of moles of electrons transferred during the experiment. We can do this by dividing the total charge by the Faraday constant. Moles of electrons = Total charge / Faraday constant Moles of electrons = 3600 C / 96,485 C/mol = 0.0373 mol Copper sulfate (CuSO4) undergoes electrolysis to produce copper (Cu) and sulfate ions (SO4^2-). The balanced half-reaction for the reduction of copper ions at the cathode is: Cu^2+ + 2e- → Cu This means that 2 moles of electrons are required to deposit 1 mole of copper. Now, we need to find the moles of copper deposited. Moles of copper = Moles of electrons / 2 Moles of copper = 0.0373 mol / 2 = 0.01865 mol Finally, we can find the mass of copper deposited by multiplying the moles of copper by the atomic weight of copper. Mass of copper = Moles of copper × Atomic weight of copper Mass of copper = 0.01865 mol × 63.546 g/mol = 1.185 g So, the mass of copper deposited on the cathode during the electrolysis experiment is approximately 1.185 grams.